3 Marks Question

Question 513 Marks

Very-Short and Short-Answer Questions:
Solve $\frac{3}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=2$ and $\frac{9}{\text{x}+\text{y}}-\frac{4}{\text{x}-\text{y}}=1$

Answer

The given system of equations is:
$\frac{3}{\text{x+y}}+\frac{2}{\text{x}-\text{y}}=2\ ....(\text{i})$
$\frac{9}{\text{x+y}}-\frac{4}{\text{x}-\text{y}}=1\ ...(\text{ii})$
Substituting $\frac{1}{\text{x+y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}$ in $(i)$ and $(ii),$ the given equations are changed to:
$3v + 2v = 2 ...(iii)$
$9u - 4v = 1 ...(iv)$
Multiplying $(i)$ by $2$ and adding it with $(ii),$ we get
$15\text{u}=4+1$
$\Rightarrow\text{u}=\frac{1}{3}$
Multiplying $(i)$ by $3$ and subtracting $(ii)$ from it, we get
$6\text{v}+4\text{v}=6-1$
$\Rightarrow\text{u}=\frac{5}{10}=\frac{1}{2}$
Therefore,
$x + y = 3 ...(v)$
$x - y = 2 ....(vi)$
Now, adding $(v)$ and $(vi)$ we have
$2\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{2}$
Substituting $\text{x}=\frac{5}{2}$ in $(v),$ we have
$\frac{5}{2}+\text{y}=3$
$\Rightarrow\text{y}=3-\frac{5}{2}=\frac{1}{2}$
Hence, $\text{x}=\frac{5}{2}$ and $\text{y}=\frac{1}{2}.$

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Question 523 Marks

Find the value of k for which the following systems of equations has unique solution:
$5x - 7y - 5 = 0, 2x + ky - 1 = 0$

Answer

$5x - 7y - 5 = 0, 2x + ky - 1 = 0$
We know that, the system of linear equations $a_1 x+b_1 y+c_1=0, a_2 x+b_2 y+c_2=0$
Has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{5}{2}\neq\frac{-7}{\text{k}}$
$\Rightarrow\ \text{k}\neq\frac{-14}{5}$

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Question 533 Marks

Very-Short and Short-Answer Questions:
Write the value of k for which the system of equations $3x + ky = 0, 2x - y = 0$ has a unique solution.

Answer

The given pair of linear equation is:
$3x + ky = 0 ...(i)$
$2x - y = 0 ...(ii)$
Which is of the form $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$,
where $a_1=3, b_1=k, c_1=0, a_2=2, b_2=-1$ and $c_2=0$
For the system to have a unique solution, we must have
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}$
$\frac{3}{2}\neq\frac{\text{k}}{-1}$
$\Rightarrow\text{k}\neq-\frac{3}{2}$
Hence, $\text{k}\neq-\frac{3}{2}.$

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Question 543 Marks

Solve for $x$ and $y:$
$\text{2x}-\frac{\text{3y}}4{}=3,$
$5\text{x}=2\text{y}+7$

Answer

The given equations are:
$\text{2x}-\frac{\text{3y}}{4}=3\ \dots(1)$
$\text{5x}=\text{2y}+7\ \dots(2)$
Multiply $(1)$ by $2$ and $2$ by $\frac{3}{4}$
$\text{4x}-\frac{\text{3y}}{2}=6\ \dots(3)$
$\frac{15}{4}\text{x}-\frac{3}{2}\text{y}-\frac{21}{4}\ \dots(4)$
Subtracting $(3)$ from $(4),$ we get
$-\frac{1}{4}\text{x}=-\frac{3}4{}$
$-\text{x}=-3$
$\Rightarrow\text{x}=3$
Substitution $x = 3$ in $(1)$, we get
$2\times3-\frac{\text{3y}}{4}=3$
$-\frac{\text{3y}}{4}=3-6$
$-\frac{\text{3y}}{4}=-3$
$\Rightarrow\text{y}=\frac{-3\times4}{-3}=4$
$\therefore$ solution is $x = 3$ and $y = 4$

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Maths 3 Marks Question